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by Jim@StratusPrep » Tue Feb 26, 2013 10:09 am
(Largest number in sequence that is a multiple of 3) - (number in sequence that is a multiple of 3) / Space between each multiple + 1

(99 - -99)/3 + 1


198/3 + 1



66 + 1 = 67


Let me know if this makes sense...
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by Anurag@Gurome » Tue Feb 26, 2013 9:47 pm
gaurav7infy wrote:how many multiples of 3 are present between -100 and 100?
Number of multiples of 3 between -100 and 100 = (Number of multiples of 3 between -1 and -100 + Number of multiples of 3 between -1 and -100) + 1 (for zero)

Now, number of multiples of 3 between 1 and 100 = Number of multiples of 3 between -1 and -100

Number of multiples of 3 between 1 and 100 = (100 - 1)/3 = 99/3 = 33

Hence, Number of multiples of 3 between -100 and 100 = 33 + 33 + 1 = 67
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by Brent@GMATPrepNow » Wed Feb 27, 2013 7:34 am
gaurav7infy wrote:how many multiples of 3 are present between -100 and 100?
Here's an alternative approach.

For this approach, all you need is one rule: the number of integers from x to y inclusive equals y - x + 1

We're looking for the number of integers in the set {-99, -96, -93, . . . . 93, 96, 99}
Since they're all multiples of 3, we can write them as:
-99 = 3(-33)
-96 = 3(-32)
-93 = 3(-31)
.
.
.
93 = 3(31)
96 = 3(32)
99 = 3(33)

At this point, all we need to do is determine the number of integers from -33 to 33 inclusive.
Applying our rule, this equals 33 - (-33) + 1 = 67

Cheers,
Brent
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